Q.Given $A = \begin{bmatrix} \sqrt{3} & 1 & -1 \\ 2 & 3 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & \sqrt{5} & 1 \\ -2 & 3 & \frac{1}{2} \end{bmatrix}$, find $A + B$.

Concept: Matrix Addition — add corresponding entries.

Step 1: Check dimensions. Both $A$ and $B$ are $2 \times 3$, so addition is defined.

Step 2: Add entry-wise:

• Row 1: $(\sqrt{3}+2,\; 1+\sqrt{5},\; -1+1)$
• Row 2: $(2+(-2),\; 3+3,\; 0+\frac{1}{2})$

Step 3: Simplify:

• Row 1: $(\sqrt{3}+2,\; 1+\sqrt{5},\; 0)$ …

Why Matrix Addition Works This Way

Matrix addition is the simplest operation in linear algebra, and it works exactly like adding two tables of numbers. If you think of a matrix as a rectangular grid of numbers, then adding two matrices means adding the number in the top-left corner of the first to the top-left corner of the second, and so on for every position.

The only rule is that both matrices must have the same shape — same number of rows and same number of columns. Here, both $A$ and $B$ are $2 \times 3$ matrices (2 rows, 3 columns), so we can add them directly.

Step-by-Step Solution

1. Identify the positions.

A $2 \times 3$ matrix has entries $a_{ij}$ where $i$ is the row number (1 or 2) and $j$ is the column number (1, 2, or 3). We'll add corresponding entries.

2. First row, first column.

$A_{11} = \sqrt{3}$, $B_{11} = 2$.

Sum: $\sqrt{3} + 2$.

3. First row, second column.

$A_{12} = 1$, $B_{12} = \sqrt{5}$.

Sum: $1 + \sqrt{5}$.

4. First row, third column.

$A_{13} = -1$, $B_{13} = 1$.

Sum: $-1 + 1 = 0$.

5. Second row, first column.

$A_{21} = 2$, $B_{21} = -2$.

Sum: $2 + (-2) = 0$. …

Method: Adding two matrices entry-by-entry

Use this for any sum $A + B$ of two matrices of the same order.

Steps

Step 1: Check the orders match.

Addition is defined only when $A$ and $B$ have the same number of rows and columns.

Step 2: Add corresponding entries.

$(A+B)_{ij} = A_{ij} + B_{ij}$; work position by position. …

Common Mistakes

Mistake 1: "Simplifying" unlike surds.

Why it's wrong: $\sqrt3 + 2$ is not $5$, and $1 + \sqrt5$ is not $\sqrt6$ — a rational and an irrational term cannot be merged. Correct approach: leave such entries as an indicated sum.

Mistake 2: Adding matrices of different orders. …